Question

Kerosene oil is flowing through a pipe having constant cross section area with diameter $2.5\text{cm}$, at a rate of $10\text{litre/sec}$. The height of bottom and top end is $2\text{m}$ and $8\text{m}$ respectively from the datum line. Find the pressure at the lower end, if the pressure at the other end is $25\times {10}^4{\text{N/m}}^2$. (Density of kerosene $=850{\text{kg/m}}^3)$

• A. $2.51\times {10}^4{\text{N/m}}^2$
• B. $1.51\times {10}^5{\text{N/m}}^2$
• C. $2\times {10}^5{\text{N/m}}^2$
• D. $3.01\times {10}^5{\text{N/m}}^2$

Answer 1

The correct option is D $3.01\times {10}^5{\text{N/m}}^2$

Let the velocity and pressure at the lower end are $v_1$ and $P_1$ respectivley.
Let the velocity and pressure at the higher level are $v_2$ and $P_2$ respectively.
Given, flow of oil $Q=10\text{litre/sec}$ or $10\times {10}^{-3}{\text{m}}^3/\text{sec}$
Using continuity equation,
$A_1{v}_1=A_1{v}_2$
but $A_1=A_2$ (uniform cross-section of tube)
$\Rightarrow v_1=v_2$
According to the Bernoulli's theorem,
$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$
$\Rightarrow P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)+\rho g(h_2-h_1)$
$[\because v_1=v_2]$
$\Rightarrow P_1-P_2=\rho g(h_2-h_1)$
$\Rightarrow P_1-25\times {10}^4=850\times 10\times (8-2)$
$\Rightarrow P_1=301000{\text{N/m}}^2$
$\Rightarrow P_1=3.01\times {10}^5{\text{N/m}}^2$